CN106354916A - Method for determining support radius and tool quantity of full face rock tunnel boring machine cutter head - Google Patents

Abstract

The invention discloses a method for determining support radius and tool quantity of a full face rock tunnel boring machine cutter head and belongs to tools for digging water conservancy and hydro-power tunnels, railway and road tunnels and urban subways. Equivalent loads of the cutter head in tunneling operation are uniformly distributed and supported on an elastic sheet on an anti-thrust bearing; according to an elastic mechanics theory, a differential equation of an elastic curved surface bent by a thin circular plate, and support relation of cutter head surface loads and the cutter head, the quantity of cutters in the cutter head support radius and the quantity of cutters beyond the support radius are determined by analyzing partial cutter head boundary conditions in the bearing and internal bearing force on the boundary; under the same quantity of cutters, the cutter head has better operating efficiency. In an embodiment, by the method, when the full face rock tunnel boring machine breaks rocks, the cutter head warping degree is reduced approximately by 25%, the amplitude and the frequency are respectively declined to 27% and 23%, and the service life of the large bearing of the cutter is increased approximately by 13%.

Description

Full face rock tunnel boring machine cutter head support radius and the determination method of number of cutters

Technical field

The invention belongs to road and bridge, culvert digging tool scope, particularly to a kind of full face rock tunnel boring machine cutter head support The determination method of radius and number of cutters.

Background technology

Cutterhead is the key core part on full face rock tunnel boring machine, and the transaction capabilities of full face rock tunnel boring machine are had Material impact, especially cutter head support radius setting with cutterhead on disk cutter quantitative relation, to cutterhead warpage, vibration and its Big bearing life has significant impact, does not still have effective workaround at present.

Content of the invention

The purpose of the present invention is to propose to a kind of full face rock tunnel boring machine knife disc tool quantity is determined with cutter head support radius Method, has been evenly arranged disk cutter before described full face rock tunnel boring machine cutterhead, supported back on big bearing, therefore, Cutterhead in excavation operation is equivalent to load uniformly and is supported on the elastic sheet on thrust bearing；It is characterized in that, comprising:

1) calculating of cutterhead load collection degree,

According to theory of elastic mechanics, the differential equation of elastic curved surface of circular sheet bending is

$(\frac{\∂2}{\∂r^2}+\frac{1}{r}\frac{\∂}{\∂r}+\frac{1}{r^2}\frac{{\∂}^2}{\∂{\mathrm{\θ}}^2})(\frac{{\∂}^{2}w}{\∂r^2}+\frac{1}{r}\frac{\∂w}{\∂r}+\frac{1}{r^2}\frac{{\∂}^{2}w}{\∂{\mathrm{\θ}}^2})=\frac{q(r,\mathrm{\θ})}{d}---\left(1\right)$

In formula: r, the θ polar coordinate with cutter head center as zero；

The amount of deflection that on w circular sheet, coordinate is put for (r, θ)；

D cutterhead bending stiffness, andWherein, e is the Young's modulus of elasticity of circular sheet material, t For lamella thickness, μ is its Poisson's ratio；

The load collection degree that q (r, θ) is distributed perpendicular to circular sheet；

Due in full face rock tunnel boring machine excavation operation, on its cutterhead every disk cutter be subject to perpendicular to cutterhead The active force in face is all very big, once the rated load of 17 inches of disk cutters is reached 250kn；For making operation cutterhead laterally not carried Lotus and tilting moment effect, the disk cutter installed thereon is designed to be uniformly and symmetrically distributed, that is, full face rock tunnel boring machine During excavation operation, the load collection degree on its cutterhead thinks uniform and equal, then

$q(r,\mathrm{\θ})=\frac{{\mathrm{nf}}_v}{{\mathrm{\πr}}^2}=q---\left(2\right)$

In formula: the disk cutter quantity installed on n cutterhead；

f_vThe normal thrust of every disk cutter；

R cutter radius；

Load collection degree meansigma methodss on q cutterhead；

Then, formula (1) is represented by:That is:

$\frac{d^{4}w}{{\mathrm{dr}}^4}+\frac{2}{r}\frac{d^{3}w}{{\mathrm{dr}}^3}-\frac{1}{r^2}\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r^3}\frac{dw}{dr}=\frac{q}{d}---\left(3\right)$

Formula (3) is a quadravalence variable coefficient nonhomogeneous linear differential equation, and its solution is expressed as particular solution w of this equation₁ With corresponding homogeneous equation

$\frac{d^{4}w}{{\mathrm{dr}}^4}+\frac{2}{r}\frac{d^{3}w}{{\mathrm{dr}}^3}-\frac{1}{r^2}\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r^3}\frac{dw}{dr}=0---\left(4\right)$

General solution w₂The form of sum；

If r=e^x, substitute into formula (4), obtain

$\frac{d^{4}w}{{\mathrm{dx}}^4}-4\frac{d^{2}w}{{\mathrm{dx}}^3}+4\frac{d^{2}w}{{\mathrm{dr}}^2}=0---\left(5\right)$

If the solution of equation (5) is w=e^kx(k is constant), substitutes into formula (5) and arranges, obtain

k⁴-4k³+4k²=0 (6)

Then, k₁=k₂=0, k₃=k₄=2；General solution w of the differential equation (5)₂It is expressed as:

w₂=(a₁+a₂x)e^0x+(a₃+a₄x)e^2x(7)

A in formula₁、a₂、a₃And a₄It is the constant being determined by boundary condition；

By r=e^xOr x=lnr substitutes into formula (7), obtain

w₂=a₁+a₂lnr+a₃r²+a₄r²lnr (8)

This is the general solution of equation (4)；

According to above-mentioned analysis, the load on cutterhead be believed that be uniformly distributed and intensity be q, therefore, the particular solution of equation (3) can It is set to w₁=cr⁴(wherein, c is constant) simultaneously substitutes into equation (3), tries to achieveThen general solution w obtaining equation (3) is:

$w=w_1+w_2=a_1+a_2\ln r+a_3{r}^2+a_4{r}^2\ln r+\frac{{\mathrm{qr}}^4}{64d}---\left(9\right)$

2) cutterhead face load and cutter head support,

(1) the face load distribution on cutterhead,

Full face rock tunnel boring machine cutterhead is typically arranged on its interior framework by the big bearing of cutterhead；Cutterhead and its big bearing Inner ring (or outer ring) is by bolted on connection, it is therefore contemplated that cutterhead is folder in its big bearing junction, and cutterhead is divided into by big bearing Two parts, can be referred to as cutterhead and divide in Bearing inner and divide in Bearing outer with cutterhead.If cutterhead upper bearing (metal) interior part arranges n_n Disk cutter, n is put in Bearing outer distribution_wDisk cutter；Set cutter head support bearing radius again as r₀, cutter radius (excavate half Footpath) it is r, according to formula (2), then the cutterhead load collection degree q that Bearing inner divides_nThe cutterhead load collection degree q dividing with Bearing outer_wRespectively For:

$\left\{\begin{array}{c}{q}_n=\frac{n_n{f}_v}{{\mathrm{\πr}}_0^2}\\ q_w=\frac{n_w{f}_v}{\mathrm{\π}(r^2-r_0^2)}\end{array}\right.---\left(10\right)$

This is the load distribution of cutterhead face；

(2) cutterhead divides boundary condition and borderline internal force in Bearing inner,

Cutterhead divides the circular sheet being regarded as central imperforate in Bearing inner, then the coefficient a in formula (9)₂And a₄Must be Zero, otherwise at cutter head center (r=0), cutterhead amount of deflection and internal force are infinity；Then the bending that divides in Bearing inner of cutterhead is scratched Spending equation is

$w_n=a_1+a_3{r}^2+\frac{q_n{r}^4}{64d}---\left(11\right)$

If the support radius of cutterhead bearing is r₀, according to folder boundary condition, have Substitution formula (11):Thus:And substitute into formula (11) arrangement, :

$w_n=\frac{q_n{r}_0^4}{64d}{(1-\frac{r^2}{r_0^2})}^2---\left(12\right)$

Thus, obtaining borderline internal force is

$\left\{\begin{array}{c}{m}_{rn}{|}_{r=r_0}=-\frac{q_n{r}_0^2}{8}\\ m_{\mathrm{\θ}n}{|}_{r=r_0}=-\frac{{\mathrm{\μq}}_n{r}_0^2}{8}\\ q_{rn}{|}_{r=r_0}=-\frac{q_n{r}_0}{2}\end{array}\right.---\left(13\right)$

(3) cutterhead divides boundary condition and borderline internal force in Bearing outer

Cutterhead divides shown in sag such as formula (9) in Bearing outer, then:

$\frac{dw}{dr}=\frac{a_2}{r}+2a_{3}r+2a_{4}r\ln r+a_{4}r+\frac{q_w{r}^2}{16d}---\left(14\right)$

$\frac{d^{2}w}{{\mathrm{dr}}^2}=-\frac{a_2}{r^2}+2a_2+3a_4+2a_4\ln r+\frac{3q_w{r}^2}{16d}---\left(15\right)$

${\▿}^{2}w=\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r}\frac{dw}{dr}=4a_2+4a_4+4a_4\ln r+\frac{q_w{r}^2}{4d}---\left(16\right)$

Its boundary condition is: in r=r₀, cutterhead folder, haveCutterhead external boundary r =r freely, has

$v_r{|}_{r=r}={(q_r+\frac{1}{r}\frac{\∂m_{r\mathrm{\θ}}}{\∂\mathrm{\θ}})}_{r=r}={\left(q_r\right)}_{r=r}={(-d\frac{d}{dr}{\▿}^{2}w)}_{r=r}=0,$

Formula (9), (14), (15) and (16) is substituted into boundary condition and solves, obtain

$\left\{\begin{array}{c}{a}_1=\frac{8q_w{r}^2{r}_0^2{\mathrm{lnr}}_0-q_w{r}_0^4}{64d}-a_2{\mathrm{lnr}}_0-a_3{r}_0^2\\ a_2=\frac{2(2r_0^2{\mathrm{lnr}}_0+r_0^2)q_w{r}^2-q_w{r}_0^4}{16d}-\frac{(3+\mathrm{\μ}+{\mathrm{lnr}}^{4+4\mathrm{\μ}})q_w{r}^4{r}_0^2+(\mathrm{\μ}-1)q_w{r}_0^3-2(\mathrm{\μ}-1)(2r_0^4{\mathrm{lnr}}_0+r_0^4)q_w{r}^2}{8d\[2(1+\mathrm{\μ})r^2-2(\mathrm{\μ}-1)r_0^2\]}\\ a_3=\frac{(3+\mathrm{\μ}+{\mathrm{lnr}}^{4+4\mathrm{\μ}})q_w{r}^4+(\mathrm{\μ}-1)q_w{r}_0^4-2(\mathrm{\μ}-1)(2r_0^2{\mathrm{lnr}}_0+r_0^2)q_w{r}^2}{16d\[2(1+\mathrm{\μ})r^2-2(\mathrm{\μ}-1)r_0^2\]}\\ a_4=-\frac{q_w{r}^2}{8d}\end{array}\right.---\left(17\right)$

Then:

$\left\{\begin{array}{c}{m}_r=-d(\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{\mathrm{\μ}}{r}\frac{dw}{dr})\\ m_{\mathrm{\θ}}=-d(\mathrm{\μ}\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r}\frac{dw}{dr})\end{array}\right.---\left(18\right)$

Formula (14), (15) are substituted into formula (18), obtains

$\left\{\begin{array}{c}{m}_r=-d\[\frac{\mathrm{\μ}-1}{r^2}{a}_2+2(1+\mathrm{\μ})a_3+(3+\mathrm{\μ}+{\mathrm{lnr}}^{2+2\mathrm{\μ}})a_4+\frac{3+\mathrm{\μ}}{16d}{q}_w{r}^2\]\\ m_{\mathrm{\θ}}=-d\[\frac{1-\mathrm{\μ}}{r^2}{a}_2+2(1+\mathrm{\μ})a_3+(3\mathrm{\μ}+1+{\mathrm{lnr}}^{2+2\mathrm{\μ}})a_4+\frac{3\mathrm{\μ}+1}{16d}{q}_w{r}^2\]\end{array}\right.---\left(19\right)$

To bearing, outer cutterhead part is made even weighing apparatus, obtains:

$2{\mathrm{\πr}}_0{q}_{r_0}+\mathrm{\π}(r^2-r_0^2)q_w=0$

So,

$q_{r_0}=-\frac{(r^2-r_0^2)q_w}{2r_0}---\left(20\right)$

In formula:Cutterhead big bearings boundary and parallel to the shearing on the cutterhead of cutterhead rotation axiss；

Other symbolic significances are the same.

3) to determine the number of cutters inside and outside cutter head support radius by the borderline internal force analysis of cutter head support,

According to above-mentioned analysis, support border to be the external boundary that Bearing inner divides, and be the inner boundary that Bearing outer divides.By In Bearing inner divide cutterhead and Bearing outer to divide cutterhead be with cutter head support as boundary demarcation, therefore, on cutter head support border On, cutterhead internal force that Bearing inner divides and the cutterhead internal force that Bearing outer divides should distinguish correspondent equal；

By in formula (20) and formula (13)Result compare, obtain

$\frac{q_n}{q_w}=\frac{r^2}{r_0^2}-1---\left(21\right)$

Formula (21), (17) are substituted into formula (19), and make r=r₀, compared with formula (13) and arrange, obtain:

$\left\{\begin{array}{c}(4\mathrm{\μ}+4)\frac{r^4}{r_0^4}ln\frac{r}{r_0}-4\mathrm{\μ}\frac{r^4}{r_0^4}-2\frac{r^4}{r_0^4}+4\mathrm{\μ}\frac{r^2}{r_0^2}-2\mathrm{\μ}+2=0\\ (4{\mathrm{\μ}}^2+4\mathrm{\μ})\frac{r^4}{r_0^4}ln\frac{r}{r_0}-4{\mathrm{\μ}}^2\frac{r^4}{r_0^4}-2\mathrm{\μ}\frac{r^4}{r_0^4}+6{\mathrm{\μ}}^2\frac{r^2}{r_0^2}-2{\mathrm{\μ}}^2+2\mathrm{\μ}=0\end{array}\right.---\left(22\right)$

Formula (22) is difficult to solve, and its value is difficult to be equal to zero, and the value that we will be equal to zero is referred to as internal force bias equivalent, point It is not designated asWithObviously, internal force bias equivalent is big, and internal force bias is also big, for this reason, formula (22) is rewritten as:

$\left\{\begin{array}{c}{\mathrm{\δ}}_{{\mathrm{rr}}_0}=(4\mathrm{\μ}+4)\frac{r^4}{r_0^4}ln\frac{r}{r_0}-4\mathrm{\μ}\frac{r^4}{r_0^4}-2\frac{r^4}{r_0^4}+4\mathrm{\μ}\frac{r^2}{r_0^2}-2\mathrm{\μ}+2\\ {\mathrm{\δ}}_{{\mathrm{\θr}}_0}=(4{\mathrm{\μ}}^2+4\mathrm{\μ})\frac{r^4}{r_0^4}ln\frac{r}{r_0}-4{\mathrm{\μ}}^2\frac{r^4}{r_0^4}-2\mathrm{\μ}\frac{r^4}{r_0^4}+6{\mathrm{\μ}}^2\frac{r^2}{r_0^2}-2{\mathrm{\μ}}^2+2\mathrm{\μ}\end{array}\right.---\left(23\right)$

By formula (10), obtain

$\frac{q_n}{q_w}=\frac{n_n}{n_w}(\frac{r^2}{r_0^2}-1)---\left(24\right)$

Compared with formula (21), obtain

$\frac{n_n}{n_w}=1---\left(25\right)$

Understand, when cutter radius are 1.62～1.63 times of its big bearings radius, and cutterhead props up in conjunction with above-mentioned analysis When number of cutters in support radius is equal with the number of cutters outside cutter head support radius, cutterhead has more preferable operational efficiency.

The invention has the beneficial effects as follows by deep theory, simulation and experimental study it was found that tunneling boring rock out Slow down its cutterhead warpage, vibration and the method improving its big bearing life during machine operation.Measured by embodiment, using this Method, can make in full face rock tunnel boring machine broken rock operation process, and cutterhead angularity reduces nearly 25%, amplitude and frequency respectively under Fall 27% and 23%, the big bearing life of cutterhead improves nearly 13%.

Brief description

Fig. 1 is to support internal force bias variation tendency on border.

Specific embodiment

The present invention proposes a kind of full face rock tunnel boring machine knife disc tool quantity and determines method with cutter head support radius, below Be explained before described full face rock tunnel boring machine cutterhead and be evenly arranged disk cutter in conjunction with accompanying drawing, embodiment, after Support, on big bearing, therefore, the cutterhead in excavation operation is equivalent to load uniformly and is supported on the Thin Elastic on thrust bearing Plate；Including:

1) calculating of cutterhead load collection degree,

According to theory of elastic mechanics, the differential equation of elastic curved surface of circular sheet bending is

$(\frac{\∂2}{\∂r^2}+\frac{1}{r}\frac{\∂}{\∂r}+\frac{1}{r^2}\frac{{\∂}^2}{\∂{\mathrm{\θ}}^2})(\frac{{\∂}^{2}w}{\∂r^2}+\frac{1}{r}\frac{\∂w}{\∂r}+\frac{1}{r^2}\frac{{\∂}^{2}w}{\∂{\mathrm{\θ}}^2})=\frac{q(r,\mathrm{\θ})}{d}---\left(1\right)$

In formula: r, the θ polar coordinate with cutter head center as zero；

The amount of deflection that on w circular sheet, coordinate is put for (r, θ)；

D cutterhead bending stiffness, andWherein, e is the Young's modulus of elasticity of circular sheet material, t For lamella thickness, μ is its Poisson's ratio；

The load collection degree that q (r, θ) is distributed perpendicular to circular sheet；

Due in full face rock tunnel boring machine excavation operation, on its cutterhead every disk cutter be subject to perpendicular to cutterhead The active force in face is all very big, once the rated load of 17 inches of disk cutters is reached 250kn；For making operation cutterhead laterally not carried Lotus and tilting moment effect, the disk cutter installed thereon is designed to be uniformly and symmetrically distributed, that is, full face rock tunnel boring machine During excavation operation, the load collection degree on its cutterhead thinks uniform and equal, then

$q(r,\mathrm{\θ})=\frac{{\mathrm{nf}}_v}{{\mathrm{\πr}}^2}=q---\left(2\right)$

In formula: the disk cutter quantity installed on n cutterhead；

f_vThe normal thrust of every disk cutter；

R cutter radius；

Load collection degree meansigma methodss on q cutterhead；

Then, formula (1) is represented by:That is:

$\frac{d^{4}w}{{\mathrm{dr}}^4}+\frac{2}{r}\frac{d^{3}w}{{\mathrm{dr}}^3}-\frac{1}{r^2}\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r^3}\frac{dw}{dr}=\frac{q}{d}---\left(3\right)$

Formula (3) is a quadravalence variable coefficient nonhomogeneous linear differential equation, and its solution is expressed as particular solution w of this equation₁ With corresponding homogeneous equation

$\frac{d^{4}w}{{\mathrm{dr}}^4}+\frac{2}{r}\frac{d^{3}w}{{\mathrm{dr}}^3}-\frac{1}{r^2}\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r^3}\frac{dw}{dr}=0---\left(4\right)$

General solution w₂The form of sum；

If r=e^x, substitute into formula (4), obtain

$\frac{d^{4}w}{{\mathrm{dx}}^4}-4\frac{d^{2}w}{{\mathrm{ax}}^3}+4\frac{d^{2}w}{{\mathrm{dr}}^2}=0---\left(5\right)$

If the solution of equation (5) is w=e^kx(k is constant), substitutes into formula (5) and arranges, obtain

k⁴-4k³+4k²=0 (6)

Then, k₁=k₂=0, k₃=k₄=2；General solution w of the differential equation (5)₂It is expressed as:

w₂=(a₁+a₂x)e^0x+(a₃+a₄x)e^2x(7)

A in formula₁、a₂、a₃And a₄It is the constant being determined by boundary condition；

By r=e^xOr x=lnr substitutes into formula (7), obtain

w₂=a₁+a₂lnr+a₃r²+a₄r²lnr (8)

This is the general solution of equation (4)；

According to above-mentioned analysis, the load on cutterhead be believed that be uniformly distributed and intensity be q, therefore, the particular solution of equation (3) can It is set to w₁=cr⁴(wherein, c is constant) simultaneously substitutes into equation (3), tries to achieveThen general solution w obtaining equation (3) is:

$w=w_1+w_2=a_1+a_2\ln r+a_3{r}^2+a_4{r}^2\ln r+\frac{{\mathrm{qr}}^4}{64d}---\left(9\right)$

2) cutterhead face load and cutter head support,

(1) the face load distribution on cutterhead,

Full face rock tunnel boring machine cutterhead is typically arranged on its interior framework by the big bearing of cutterhead；Cutterhead and its big bearing Inner ring (or outer ring) is by bolted on connection, it is therefore contemplated that cutterhead is folder in its big bearing junction, and cutterhead is divided into by big bearing Two parts, can be referred to as cutterhead and divide in Bearing inner and divide in Bearing outer with cutterhead.If cutterhead upper bearing (metal) interior part arranges n_n Disk cutter, n is put in Bearing outer distribution_wDisk cutter；Set cutter head support bearing radius again as r₀, cutter radius (excavate half Footpath) it is r, according to formula (2), then the cutterhead load collection degree q that Bearing inner divides_nThe cutterhead load collection degree q dividing with Bearing outer_wRespectively For:

$\left\{\begin{array}{c}{q}_n=\frac{n_n{f}_v}{{\mathrm{\πr}}_0^2}\\ q_w=\frac{n_w{f}_v}{\mathrm{\π}(r^2-r_0^2)}\end{array}\right.---\left(10\right)$

This is the load distribution of cutterhead face；

(2) cutterhead divides boundary condition and borderline internal force in Bearing inner,

Cutterhead divides the circular sheet being regarded as central imperforate in Bearing inner, then the coefficient a in formula (9)₂And a₄Must be Zero, otherwise at cutter head center (r=0), cutterhead amount of deflection and internal force are infinity.Then the bending that divides in Bearing inner of cutterhead is scratched Spending equation is

$w_n=a_1+a_3{r}^2+\frac{q_n{r}^4}{64d}---\left(11\right)$

If the support radius of cutterhead bearing is r₀, according to folder boundary condition, have Substitution formula (11):Thus:And substitute into formula (11) arrangement, :

$w_n=\frac{q_n{r}_0^4}{64d}{(1-\frac{r^2}{r_0^2})}^2---\left(12\right)$

Thus, obtaining borderline internal force is

$\left\{\begin{array}{c}{m}_{rn}{|}_{r=r_0}=-\frac{q_n{r}_0^2}{8}\\ m_{\mathrm{\θ}n}{|}_{r=r_0}=-\frac{{\mathrm{\μq}}_n{r}_0^2}{8}\\ q_{rn}{|}_{r=r_0}=-\frac{q_n{r}_0}{2}\end{array}\right.---\left(13\right)$

(3) cutterhead divides boundary condition and borderline internal force in Bearing outer

Cutterhead divides shown in sag such as formula (9) in Bearing outer, then:

$\frac{dw}{dr}=\frac{a_2}{r}+2a_{3}r+2a_{4}r\ln r+a_{4}r+\frac{q_w{r}^2}{16d}---\left(14\right)$

$\frac{d^{2}w}{{\mathrm{dr}}^2}=-\frac{a_2}{r^2}+2a_3+3a_4+2a_4\ln r+\frac{3q_w{r}^2}{16d}---\left(15\right)$

${\▿}^{2}w=\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r}\frac{dw}{dr}=4a_3+4a_4+4a_4\ln r+\frac{q_w{r}^2}{4d}---\left(16\right)$

Its boundary condition is: in r=r₀, cutterhead folder, haveCutterhead external boundary r =r freely, has

$v_r{|}_{r=r}={(q_r+\frac{1}{r}\frac{\∂m_{r\mathrm{\θ}}}{\∂\mathrm{\θ}})}_{r=r}={\left(q_r\right)}_{r=r}={(-d\frac{d}{dr}{\▿}^{2}w)}_{r=r}=0,$

Formula (9), (14), (15) and (16) is substituted into boundary condition and solves, obtain

$\left\{\begin{array}{c}{a}_2=\frac{8q_w{r}^2{r}_0^2{\mathrm{lnr}}_0-q_w{r}_0^4}{64d}-a_2{\mathrm{lnr}}_0-a_3{r}_0^2\\ a_2=\frac{2(2r_0^2{\mathrm{lnr}}_0+r_0^2)q_w{r}^2-q_w{r}_0^4}{16d}-\frac{(3+\mathrm{\μ}+{\mathrm{lnr}}^{4+4\mathrm{\μ}})q_w{r}^4{r}_0^2+(\mathrm{\μ}-1)q_w{r}_0^3-2(\mathrm{\μ}-1)(2r_0^4{\mathrm{lnr}}_0+r_0^4)q_w{r}^2}{8d\[2(1+\mathrm{\μ})r^2-2(\mathrm{\μ}-1)r_0^2\]}\\ a_3=\frac{(3+\mathrm{\μ}+{\mathrm{lnr}}^{4+4\mathrm{\μ}})q_w{r}^4+(\mathrm{\μ}-1)q_w{r}_0^4-2(\mathrm{\μ}-1)(2r_0^2{\mathrm{lnr}}_0+r_0^2)q_w{r}^2}{16d\[2(1+\mathrm{\μ})r^2-2(\mathrm{\μ}-1)r_0^2\]}\\ a_4=-\frac{q_w{r}^2}{8d}\end{array}\right.---\left(17\right)$

Then:

$\left\{\begin{array}{c}{m}_r=-d(\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{\mathrm{\μ}}{r}\frac{dw}{dr})\\ m_{\mathrm{\θ}}=-d(\mathrm{\μ}\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r}\frac{dw}{dr})\end{array}\right.---\left(18\right)$

Formula (14), (15) are substituted into formula (18), obtains

$\left\{\begin{array}{c}{m}_r=-d\[\frac{\mathrm{\μ}-1}{r^2}{a}_2+2(1+\mathrm{\μ})a_3+(3+\mathrm{\μ}+{\mathrm{lnr}}^{2+2\mathrm{\μ}})a_4+\frac{3+\mathrm{\μ}}{16d}{q}_w{r}^2\]\\ m_{\mathrm{\θ}}=-d\[\frac{1-\mathrm{\μ}}{r^2}{a}_2+2(1+\mathrm{\μ})a_3+(3\mathrm{\μ}+1+{\mathrm{lnr}}^{2+2\mathrm{\μ}})a_4+\frac{3\mathrm{\μ}+1}{16d}{q}_w{r}^2\]\end{array}\right.---\left(19\right)$

To bearing, outer cutterhead part is made even weighing apparatus, obtains:

$2{\mathrm{\πr}}_0{q}_{r_0}+\mathrm{\π}(r^2-r_0^2)q_w=0$

So,

$q_{r_0}=-\frac{(r^2-r_0^2)q_w}{2r_0}---\left(20\right)$

In formula:Cutterhead big bearings boundary and parallel to the shearing on the cutterhead of cutterhead rotation axiss；

Other symbolic significances are the same.

3) to determine the number of cutters inside and outside cutter head support radius by the borderline internal force analysis of cutter head support,

According to above-mentioned analysis, support border to be the external boundary that Bearing inner divides, and be the inner boundary that Bearing outer divides.By In Bearing inner divide cutterhead and Bearing outer to divide cutterhead be with cutter head support as boundary demarcation, therefore, on cutter head support border On, cutterhead internal force that Bearing inner divides and the cutterhead internal force that Bearing outer divides should distinguish correspondent equal；

By in formula (20) and formula (13)Result compare, obtain

$\frac{q_n}{q_w}=\frac{r^2}{r_0^2}-1---\left(21\right)$

Formula (21), (17) are substituted into formula (19), and make r=r₀, compared with formula (13) and arrange, obtain:

$\left\{\begin{array}{c}(4\mathrm{\μ}+4)\frac{r^4}{r_0^4}ln\frac{r}{r_0}-4\mathrm{\μ}\frac{r^4}{r_0^4}-2\frac{r^4}{r_0^4}+4\mathrm{\μ}\frac{r^2}{r_0^2}-2\mathrm{\μ}+2=0\\ (4{\mathrm{\μ}}^2+4\mathrm{\μ})\frac{r^4}{r_0^4}ln\frac{r}{r_0}-4{\mathrm{\μ}}^2\frac{r^4}{r_0^4}-2\mathrm{\μ}\frac{r^4}{r_0^4}+6{\mathrm{\μ}}^2\frac{r^2}{r_0^2}-2{\mathrm{\μ}}^2+2\mathrm{\μ}=0\end{array}\right.---\left(22\right)$

Formula (22) is difficult to solve, and its value is difficult to be equal to zero, and the value that we will be equal to zero is referred to as internal force bias equivalent, point It is not designated asWithObviously, internal force bias equivalent is big, and internal force bias is also big, for this reason, formula (22) is rewritten as:

$\left\{\begin{array}{c}{\mathrm{\δ}}_{{\mathrm{rr}}_0}=(4\mathrm{\μ}+4)\frac{r^4}{r_0^4}ln\frac{r}{r_0}-4\mathrm{\μ}\frac{r^4}{r_0^4}-2\frac{r^4}{r_0^4}+4\mathrm{\μ}\frac{r^2}{r_0^2}-2\mathrm{\μ}+2\\ {\mathrm{\δ}}_{{\mathrm{\θr}}_0}=(4{\mathrm{\μ}}^2+4\mathrm{\μ})\frac{r^4}{r_0^4}ln\frac{r}{r_0}-4{\mathrm{\μ}}^2\frac{r^4}{r_0^4}-2\mathrm{\μ}\frac{r^4}{r_0^4}+6{\mathrm{\μ}}^2\frac{r^2}{r_0^2}-2{\mathrm{\μ}}^2+2\mathrm{\μ}\end{array}\right.---\left(23\right)$

By formula (10), obtain

$\frac{q_n}{q_w}=\frac{n_n}{n_w}(\frac{r^2}{r_0^2}-1)---\left(24\right)$

Compared with formula (21), obtain

$\frac{n_n}{n_w}=1---\left(25\right)$

Understand, when cutter radius are 1.62～1.63 times of its big bearings radius, and cutterhead props up in conjunction with above-mentioned analysis When number of cutters in support radius is equal with the number of cutters outside cutter head support radius, cutterhead has more preferable operational efficiency.

By μ=0.28 and differentSubstitution formula (23), obtains cutter head support borderline internal force bias equivalent weight values solution (being shown in Table 1), the variation tendency of this bias equivalent is as shown in Figure 1.Be can be seen that in given numerical solution by table 1 and Fig. 1 In sequence, on cutter head support border, internal force bias equivalent absolute value meansigma methodss are minimum occursPlace, footpath herein Only have 1% to moment of flexure bias equivalent, circumferential moment of flexure bias equivalent is 41.1%.Circumferential moment of flexure bias is big, to cutter axis Holding capacity is unfavorable；Radial direction moment of flexure bias is big, and in full face rock tunnel boring machine construction operation, cutterhead easily occurs warpage, thus increasing Big vibration.Therefore, full face rock tunnel boring machine cutter head support is arranged onPlace, relatively rationally.Gone back by table 1 As can be seen thatOn cutter head support border internal force bias equivalent absolute value meansigma methodss all 24.5% with Under, this substantially can meet requirement of engineering.

Table 1 internal force bias equivalent

Embodiment

Known certain tunnel excavation diameter 7.66m, according to wall rock geology condition, on full face rock tunnel boring machine cutterhead Disk cutter 56 need to be set.

Then:

By patent requirements, 1) take r₀=2.36 (m), then:

2) disk cutter 28 is set in cutter head support radius, disk cutter 28 is set outside cutter head support radius.

Measured according to numerical simulation and test, in full face rock tunnel boring machine broken rock operation process, cutterhead angularity relatively passes The cutterhead of system method for designing design reduces nearly 25%, and amplitude and frequency decline 27% and 23% respectively, and the big bearing life of cutterhead carries High nearly 13%.

Claims (1)

1. a kind of full face rock tunnel boring machine knife disc tool quantity and cutter head support radius determine method, described tunneling boring rock pick Enter and before machine knife disk, be evenly arranged disk cutter, supported back on big bearing, therefore, will be equivalent for the cutterhead in excavation operation Uniformly and it is supported on the elastic sheet on thrust bearing for load；It is characterized in that, comprising:

1) calculating of cutterhead load collection degree,

According to theory of elastic mechanics, the differential equation of elastic curved surface of circular sheet bending is

$(\frac{{\∂}^2}{\∂r^2}+\frac{1}{r}\frac{\∂}{\∂r}+\frac{1}{r^2}\frac{{\∂}^2}{\∂{\mathrm{\θ}}^2})(\frac{{\∂}^{2}w}{\∂r^2}+\frac{1}{r}\frac{\∂w}{\∂r}+\frac{1}{r^2}\frac{{\∂}^{2}w}{\∂{\mathrm{\θ}}^2})=\frac{q(r,\mathrm{\θ})}{d}---\left(1\right)$

In formula: r, the θ polar coordinate with cutter head center as zero；

The amount of deflection that on w circular sheet, coordinate is put for (r, θ)；

D cutterhead bending stiffness, andWherein, e is the Young's modulus of elasticity of circular sheet material, and t is thin Plate thickness, μ is its Poisson's ratio；

The load collection degree that q (r, θ) is distributed perpendicular to circular sheet；

Due in full face rock tunnel boring machine excavation operation, on its cutterhead every disk cutter be subject to perpendicular to cutterhead face Active force is all very big, once the rated load of 17 inches of disk cutters is reached 250kn；For make operation cutterhead be not subject to transverse load and Tilting moment acts on, and the disk cutter installed thereon is designed to be uniformly and symmetrically distributed, that is, full face rock tunnel boring machine driving During operation, the load collection degree on its cutterhead thinks uniform and equal, then

$q(r,\mathrm{\θ})=\frac{{\mathrm{nf}}_v}{{\mathrm{\πr}}^2}=q---\left(2\right)$

In formula: the disk cutter quantity installed on n cutterhead；

f_vThe normal thrust of every disk cutter；

R cutter radius；

Load collection degree meansigma methodss on q cutterhead；

Then, formula (1) is represented by:That is:

$\frac{d^{4}w}{{\mathrm{dr}}^4}+\frac{2}{r}\frac{d^{3}w}{{\mathrm{dr}}^3}-\frac{1}{r^2}\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r^3}\frac{dw}{dr}=\frac{q}{d}---\left(3\right)$

Formula (3) is a quadravalence variable coefficient nonhomogeneous linear differential equation, and its solution is expressed as particular solution w of this equation₁And with its Corresponding homogeneous equation

$\frac{d^{4}w}{{\mathrm{dr}}^4}+\frac{2}{r}\frac{d^{3}w}{{\mathrm{dr}}^3}-\frac{1}{r^2}\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r^3}\frac{dw}{dr}=0---\left(4\right)$

General solution w₂The form of sum；

If r=e^x, substitute into formula (4), obtain

$\frac{d^{4}w}{{\mathrm{dx}}^4}-4\frac{d^{3}w}{{\mathrm{dx}}^3}+4\frac{d^{4}w}{{\mathrm{dr}}^2}=0---\left(5\right)$

If the solution of equation (5) is w=e^kx(k is constant), substitutes into formula (5) and arranges, obtain

k⁴-4k³+4k²=0 (6)

Then, k₁=k₂=0, k₃=k₄=2；General solution w of the differential equation (5)₂It is expressed as:

w₂=(a₁+a₂x)e^0x+(a₃+a₄x)e^2x(7)

A in formula₁、a₂、a₃And a₄It is the constant being determined by boundary condition；

By r=e^xOr x=lnr substitutes into formula (7), obtain

w₂=a₁+a₂lnr+a₃r²+a₄r²lnr (8)

This is the general solution of equation (4)；

According to above-mentioned analysis, the load on cutterhead be believed that be uniformly distributed and intensity be q, therefore, the particular solution of equation (3) can be set to w₁=cr⁴(wherein, c is constant) simultaneously substitutes into equation (3), tries to achieveThen general solution w obtaining equation (3) is:

$w=w_1+w_2=a_1+a_2\ln r+a_3{r}^2+a_4{r}^2\ln r+\frac{{\mathrm{qr}}^4}{64d}---\left(9\right)$

2) cutterhead face load and cutter head support,

(1) the face load distribution on cutterhead,

Full face rock tunnel boring machine cutterhead is typically arranged on its interior framework by the big bearing of cutterhead；Cutterhead and its big bearing inner race (or outer ring) is by bolted on connection, it is therefore contemplated that cutterhead is folder in its big bearing junction, and cutterhead is divided into two by big bearing Point, cutterhead can be referred to as divide in Bearing inner and divide in Bearing outer with cutterhead, if cutterhead upper bearing (metal) interior part arrangement n_nDisk Shape hobboing cutter, n is put in Bearing outer distribution_wDisk cutter；Set cutter head support bearing radius again as r₀, cutter radius be r, according to formula (2) the cutterhead load collection degree q that, then Bearing inner divides_nThe cutterhead load collection degree q dividing with Bearing outer_wIt is respectively as follows:

$\left\{\begin{array}{c}{q}_n=\frac{n_n{f}_v}{{\mathrm{\πr}}_0^2}\\ q_w=\frac{n_w{f}_v}{\mathrm{\π}(r^2-r_{\mathrm{\θ}}^2)}\end{array}\right.---\left(10\right)$

This is the load distribution of cutterhead face；

(2) cutterhead divides boundary condition and borderline internal force in Bearing inner,

Cutterhead divides the circular sheet being regarded as central imperforate in Bearing inner, then the coefficient a in formula (9)₂And a₄Must be zero, no Then at cutter head center (r=0), cutterhead amount of deflection and internal force are infinity；Then obtain the sag side that cutterhead divides in Bearing inner Cheng Wei

$w_n=a_1+a_3{r}^2+\frac{q_n{r}^4}{64d}---\left(11\right)$

If the support radius of cutterhead bearing is r₀, according to folder boundary condition, have Substitution formula (11):Thus:And substitute into formula (11) arrangement, :

$w_n=\frac{q_n{r}_0^4}{64d}{(1-\frac{r^2}{r_0^2})}^2---\left(12\right)$

Thus, obtaining borderline internal force is

$\left\{\begin{array}{c}{m}_{rn}{|}_{r=r_0}=-\frac{q_n{r}_0^2}{8}\\ m_{\mathrm{\θ}n}{|}_{r=r_0}=-\frac{{\mathrm{\μq}}_n{r}_0^2}{8}\\ q_{rn}{|}_{r=r_0}=-\frac{q_n{r}_0}{2}\end{array}\right.---\left(13\right)$

(3) cutterhead divides boundary condition and borderline internal force in Bearing outer

Cutterhead divides shown in sag such as formula (9) in Bearing outer, then:

$\frac{dw}{dr}=\frac{a_2}{r}+2a_{3}r+2a_{4}r\ln r+a_{4}r+\frac{q_w{r}^3}{16d}---\left(14\right)$

$\frac{d^{2}w}{{\mathrm{dr}}^2}=-\frac{a_2}{r^2}+2a_2+3a_4+2a_4\ln r+\frac{3q_w{r}^2}{16d}---\left(15\right)$

${\▿}^{2}w=\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r}\frac{dw}{dr}=4a_3+4a_4+4a_4\ln r+\frac{q_w{r}^2}{4d}---\left(16\right)$

Its boundary condition is: in r=r₀, cutterhead folder, haveCutterhead external boundary r=r Freely, have

$v_r{|}_{r=r}={(q_r+\frac{1}{r}\frac{\∂m_{r\mathrm{\θ}}}{\∂\mathrm{\θ}})}_{r=r}={\left(q_r\right)}_{r=r}={(-d\frac{d}{dr}{\▿}^{2}w)}_{r=r}=0,$

Formula (9), (14), (15) and (16) is substituted into boundary condition and solves, obtain

$\left\{\begin{array}{c}{a}_1=\frac{8q_w{r}^2{r}_0^2{\mathrm{lnr}}_0-q_w{r}_0^4}{64d}-a_2{\mathrm{lnr}}_0-a_3{r}_0^2\\ a_2=\frac{2(2r_0^2{\mathrm{lnr}}_0+r_0^2)q_w{r}^2-q_w{r}_0^4}{16d}-\frac{(3+\mathrm{\μ}+{\mathrm{lnr}}^{4+4\mathrm{\μ}})q_w{r}^4{r}_0^2+(\mathrm{\μ}-1)q_w{r}_0^6-2(\mathrm{\μ}-1)(\mathrm{\μ}-1)(2r_0^4{\mathrm{lnr}}_0+r_0^4)q_0{r}^2}{8d\[2(1+\mathrm{\μ})r^2-2(\mathrm{\μ}-1)r_0^2\]}\\ a_2=\frac{(3+\mathrm{\μ}+{\mathrm{lnr}}^{4+4\mathrm{\μ}})q_w{r}^4+(\mathrm{\μ}-1)q_w{r}_0^4-2(\mathrm{\μ}-1)(2r_0^2{\mathrm{lnr}}_0+r_0^2)q_w{r}^2}{16d\[2(1+\mathrm{\μ})r^2-2(\mathrm{\μ}-1)r_0^2\]}\\ a_4=-\frac{q_w{r}^2}{8d}\end{array}\right.---\left(17\right)$

Then:

$\left\{\begin{array}{c}{m}_r=-d(\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{\mathrm{\μ}}{r}\frac{dw}{dr})\\ m_{\mathrm{\θ}}=-d(\mathrm{\μ}\frac{d^{2}w}{{\mathrm{dr}}^2}+\frac{1}{r}\frac{dw}{dr})\end{array}\right.---\left(18\right)$

Formula (14), (15) are substituted into formula (18), obtains

$\left\{\begin{array}{c}{m}_r=-d\[\frac{\mathrm{\μ}-1}{r^2}{a}_2+2(1+\mathrm{\μ})a_3+(3+\mathrm{\μ}+{\mathrm{lnr}}^{2+2\mathrm{\μ}})a_4+\frac{3+\mathrm{\μ}}{16d}{q}_w{r}^2\]\\ m_{\mathrm{\θ}}=-d\[\frac{1-\mathrm{\μ}}{r^2}{a}_2+2(1+\mathrm{\μ})a_3+(3\mathrm{\μ}+1+{\mathrm{lnr}}^{2+2\mathrm{\μ}})a_4+\frac{3\mathrm{\μ}+1}{16d}{q}_w{r}^2\]\end{array}\right.---\left(19\right)$

To bearing, outer cutterhead part is made even weighing apparatus, obtains:

$3{\mathrm{\πr}}_0{q}_{r_0}+\mathrm{\π}(r^2-r_0^2)q_w=0$

So,

$q_{r_0}=-\frac{(r^2-r_0^2)q_w}{2r_0}---\left(20\right)$

In formula:Cutterhead big bearings boundary and parallel to the shearing on the cutterhead of cutterhead rotation axiss；

3) to determine the number of cutters inside and outside cutter head support radius by the borderline internal force analysis of cutter head support,

According to above-mentioned analysis, support border to be the external boundary that Bearing inner divides, and be the inner boundary that Bearing outer divides, due to axle Hold interior part cutterhead and Bearing outer to divide cutterhead be with cutter head support as boundary demarcation, therefore, on cutter head support border, axle Hold the cutterhead internal force of interior part and cutterhead internal force that Bearing outer divides should distinguish correspondent equal；

By in formula (20) and formula (13)Result compare, obtain

$\frac{q_n}{q_w}=\frac{r^2}{r_0^2}-1---\left(21\right)$

Formula (21), (17) are substituted into formula (19), and make r=r₀, compared with formula (13) and arrange, obtain:

$\left\{\begin{array}{c}(4\mathrm{\μ}+4)\frac{r^4}{r_0^4}\ln \frac{r}{r_0}-4\mathrm{\μ}\frac{r^4}{r_0^4}-2\frac{r^4}{r_0^4}+4\mathrm{\μ}\frac{r^2}{r_0^2}-2\mathrm{\μ}+2=0\\ (4{\mathrm{\μ}}^2+4\mathrm{\μ})\frac{r^4}{r_0^4}\ln \frac{r}{r_0}-4{\mathrm{\μ}}^2\frac{r^4}{r_0^4}-2\mathrm{\μ}\frac{r^4}{r_0^4}+6{\mathrm{\μ}}^2\frac{r^2}{r_0^2}-2{\mathrm{\μ}}^2+2\mathrm{\μ}=0\end{array}\right.---\left(22\right)$

Formula (22) is difficult to solve, and its value is difficult to be equal to zero, and the value that we will be equal to zero is referred to as internal force bias equivalent, remembers respectively ForWithObviously, internal force bias equivalent is big, and internal force bias is also big, for this reason, formula (22) is rewritten as:

$\left\{\begin{array}{c}{\mathrm{\δ}}_{{\mathrm{rr}}_0}=(4\mathrm{\μ}+4)\frac{r^4}{r_0^4}\ln \frac{r}{r_0}-4\mathrm{\μ}\frac{r^4}{r_0^4}-2\frac{r^4}{r_0^4}+4\mathrm{\μ}\frac{r^2}{r_0^2}-2\mathrm{\μ}+2\\ {\mathrm{\δ}}_{{\mathrm{\θr}}_0}=(4{\mathrm{\μ}}^2+4\mathrm{\μ})\frac{r^4}{r_0^4}\ln \frac{r}{r_0}-4{\mathrm{\μ}}^2\frac{r^4}{r_0^4}-2\mathrm{\μ}\frac{r^4}{r_0^4}+6{\mathrm{\μ}}^2\frac{r^2}{r_0^2}-2{\mathrm{\μ}}^2+2\mathrm{\μ}\end{array}\right.---\left(23\right)$

By formula (10), obtain

$\frac{q_n}{q_w}=\frac{n_n}{n_w}(\frac{r^2}{r_0^2}-1)---\left(24\right)$

Compared with formula (21), obtain

$\frac{n_n}{n_w}=1---\left(25\right)$

Understand in conjunction with above-mentioned analysis, when cutter radius are 1.62～1.63 times of its big bearings radius, and cutter head support half When number of cutters in footpath is equal with the number of cutters outside cutter head support radius, cutterhead has more preferable operational efficiency.